Question

. A discount brokerage selected a random sample of 64 customers and reviewed the value of their accounts. The mean was $32,000 with a population standard deviation of $8,200. What is a 90% confidence interval for the mean account value of the population of customers

Answer 1

Answer:

The  90% confidence interval is  $\ \ 30313.9< \mu < \ \ 33686.13$

Step-by-step explanation:

From the question we are told that

   The  sample size is  n =  64

     The sample  mean is  $\= x = \ 32, 000$

     The  standard deviation is  $\sigma= \ 8, 200$

     

Given that the confidence interval is  90% then the level of significance is mathematically evaluated as

             $\alpha = 100 - 90$

             $\alpha = 10 \%$

            $\alpha = 0.10$

Next we obtain the critical value of  $\frac{ \alpha }{2}$ from the normal distribution table , the value is  

       $Z_{\frac{\alpha }{2} } = 1.645$

  Generally the margin of error is mathematically represented as

        $E = Z_{\frac{\alpha }{2} } * \frac{ \sigma }{ \sqrt{n} }$

  =>   $E = 1.645 * \frac{ 8200 }{ \sqrt{64} }$

  =>   $E = 1686.13$

The 90% confidence interval is mathematically represented as

      $\= x - E < \mu < \= x + E$

 =>    $32000 - 1689.13 < \mu < 32000 + 1689.13$

=>    $\ \ 30313.9< \mu < \ \ 33686.13$